2 Marks Questions

Question 12 Marks

Find the centroid of a triangle, the mid-point of whose sides are D(1, 2, -3), E(3, 0, 1) and F(-1, 1, -4).

Answer

Given that, mid-points of sides of $\triangle\text{ABC}$ are D(1, 2, -3), E(3, 0, 1) and F(-1 1,-4).
Now from the geometry of centroid,
we know that the centroid of $\triangle\text{DEF}$ is
same as the centroid of $\triangle\text{ABC}.$
$\therefore$ Centroid of $\triangle\text{ABC}$ is
$\text{G}\Big(\frac{1+3-1}{3},\frac{2+0+1}{3},\frac{-3+1-4}{3}\Big)$
$\equiv\text{G}(1,1,-2)$

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Question 22 Marks

Find the distance from the origin to (6, 6, 7).

Answer

Coordinates of the origin are (0, 0, 0)
$\therefore$ Distance from (0, 0, 0) to (6, 6, 7)
$=\sqrt{(6-0)^2+(6-0)^2+(7-0)^2}$
$=\sqrt{36+36+49}=\sqrt{121}$
= 11 units.
Hence, the required distance = 11 units.

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Question 32 Marks

If the origin is the centriod of a triangle ABC having vertices A(a, 1, 3), B(-2, b, -5) and C(4, 7, c), find the values of a, b, c.

Answer

Coordinates of the centroid G = (0, 0, 0)
$\therefore0=\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3}\Rightarrow0=\frac{\text{a}-2+4}{3}\Rightarrow\text{a}=-2$
$0=\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Rightarrow0=\frac{1+\text{b}+7}{3}\Rightarrow\text{b}=-8$
and $0=\frac{\text{z}_1+\text{z}_2+\text{z}_3}{3}\Rightarrow0=\frac{3-5+\text{c}}{3}\Rightarrow\text{c}=2$
Hence, the reuired values are a = -2, b = -8 and c = 2.

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Question 42 Marks

What are the coordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with the origin and the three edges passing through the origin, coincides with the positive direction of the axes through the origin?

Answer

Given that each edge of the cuboid is 2 units.
$\therefore$ coordinates of the vertices are

A(2, 0, 0), B(2, 2, 0), C(0, 2, 0), D(0, 2, 2), E(0, 0, 2), F(2, 0, 2), G(2, 2, 2), and O(0, 0, 0).

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Question 52 Marks

Locate the following points:

(1, -1, 3).
(-1, 2, 4).
(-2, -4, -7).
(-4, 2, -5).

Answer

Given, coordinates:

(1, -1, 3).
(-1, 2, 4).
(-2, -4, -7).
(-4, 2, -5).

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Question 62 Marks

Show that if $\text{x}^2 + \text{y}^2 = 1,$ then the point $\Big(\text{x, y},\sqrt{1-\text{x}^2-\text{y}^2}\Big)$ is at a distance 1 unit from the origin.

Answer

Given that, $\text{x}^2 + \text{y}^2 = 1$
$\therefore$ Distance of the point $\Big(\text{x, y, }\sqrt{1-\text{x}^2-\text{y}^2}\Big)$ from origin is given as
$\text{d}=\sqrt{\text{x}^2+\text{y}^2+\big(\sqrt{1-\text{x}^2-\text{y}^2\big)^2}}$ $=\sqrt{\text{x}^2+\text{y}^2+1-\text{x}^2-\text{y}^2}=1$

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Question 72 Marks

Find the third vertex of triangle whose centroid is origin and two vertices are (2, 4, 6) and (0, -2, -5).

Answer

Let the third or unknown vertex of $\triangle\text{ABC}$ be A(x, y, z).
Other vertices of triangle are B(2,4, 6) and C(0, -2, -5).
The centroid is G(0, 0, 0).
$\therefore(0,0,0)=\Big(\frac{2+0+\text{x}}{3},\frac{4-2+\text{y}}{3},\frac{6-5+\text{z}}{3}\Big)$
On comparing coordinates, we get,
$\frac{2+\text{x}}{3}=0,\frac{2+\text{y}}{3}=0\ \text{and}\ \frac{1+\text{z}}{3}=0$
$\Rightarrow\text{x}=-2,\ \text{y}=-2\ \text{and}\ \text{z}=-1$

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MATHS 2 Marks Questions

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